Transcript Linear Motion

EED2023 Mechanical Principles
and Applications
WORK, POWER AND ENERGY
TRANSFER IN DYNAMIC
ENGINEERING SYSTEMS
(LINEAR MOTION)
Linear Motion
Speed, velocity and acceleration
Force and momentum
- Newton’s Laws of Motions
- Linear and angular motion
- Momentum
- Conservation of momeNtum
Speed, velocity and acceleration
 Speed – distance moved per unit second
Average speed 
distance moved
time taken
(m / s)
 Velocity – speed and direction of travel
distance moved in a particular direction AB

time taken
OB
(m/s in a particular direction – north for example)
Average velocity 
distance
distance
distance
A
B
time
Increasing velocity
O
time
time
Uniform velocity
Decreasing velocity
Speed, velocity and acceleration
 Acceleration – the change in velocity per unit second
Accelerati on 
change in velocit y
(m / s )
time taken for this change
2
velocity
velocity
A
B
time
Uniform acceleration
Acceleration 
time
Non-uniform acceleration
PQ
; thegradient of the velocity- timegraph
OQ
Uniformly Accelerated Motion
Q
Distance travelled:
velocity
Average velocity 
v
P
R
t
O
time
Acceleration, a 

and
distance moved
time taken
distance moved  average velocity x time taken
Velocity 
hence
s
u
therefore
uv
(object accelerati ng uniformly only)
2
changein velocity
time takenfor thischange
vu
t
v  u  at
v  u  at
S
u  v  x
t
(1)
2
v  u  at
but
s
therefore
(2)
u  u  at t
2
s  ut  at
or
1
2
2
(3)
elimating time t between equation (2) and (3),
we obtained :
v  u  2as
2
2
(4)
Linear Motion
 So with reference to the statement below:


Uniform velocity is motion along a straight line at constant
speed.
Uniform acceleration is motion in a straight line with constant
acceleration.
Speed, v
Time, t
 We can deduce that in a straight line from stop we get uniform
acceleration, then we get uniform velocity and then we get uniform
deceleration as depicted in the velocity graph above.
Linear Motion – Example 1
 A train accelerates uniformly from rest to reach 54km/h in
200s after which the speed remains constant for 300s, at the
end of this time the train decelerates to rest in 150s.
Determine:
- The total distance travelled
- The acceleration
- The deceleration
Answer:
Distance travelled = 7125 m / 7.125 km
Acceleration = 0.075 ms-2
Deceleration = 0.1 ms-2
Linear Motion – Example 2
 A truck accelerates uniformly from rest to a speed of
100km/h taking 2 minutes, it then remains at a constant
velocity for 32 minutes, coming up to a set of traffic lights the
truck decelerates to rest in 30s.
Determine:
- The constant velocity in m/s
- The total distance travelled
- The acceleration
- The deceleration
Answer: 27.78 m/s; 55.42 km; 0.2315 ms-2; 0.926 ms-2
Linear Motion – Example 3
 A train starting from rest leave a station with uniform
acceleration for 20s, it then proceeds at a constant velocity
before decelerating for 30s, coming to a total standstill. If the
total distance travelled is 3 km and the total time taken was 5
minutes, sketch a velocity time graph and determine:
- The uniform speed
- The acceleration
- The deceleration
- The distance travelled during the first minute
Answer: 10.91 m/s; 0.545 ms-2; 0.3637 ms-2; 545.5 m
Angular and Linear Motion
Length of a circular arc, s = R
R
where  is in radian

s
2
 2f
T
1
f 
is frequency
T
T heangular speed,  
where
For a point object moving on a circular path at a uniform speed, its speed, v, may
be calculated if the radius, R and time period, T are known. In one complete
rotation, the object travels a distance of 2R in time T. Therefore, its speed is
given by,
2R
v
T
Therefore,
v  R
Angular acceleration
When a flywheel speeds up, every point of the flywheel moves at increasing
speed. Consider a point on flywheel rim, at distance R from the axis of
rotation. If the flywheel speeds up from initial angular speed ω1 to angular
speed ω2 in time t, then the speed of the point increases from speed u = ω1 R
to speed v = ω2 R in time t. By using acceleration, a = (v-u)/t, the linear
acceleration of the point is

2  1 
a
R
t
where α is the angular acceleration of a rotating object, defined as the change
of angular velocity per second, and given by

2  1 

t
Hence, linear acceleration of a point along its circular path is given by,
a  R
Exercise
An aircraft accelerates from 100m/s to 300m/s in 100s.
What is its acceleration?
2. A car joins a motorway travelling at 15m/s. It accelerates
at 2ms-2 for 8s. How far will it travel in this time?
3. A vehicle accelerates from rest, covering 500m in 20s.
Calculate
1.
a) the acceleration
b) the velocity of the vehicle at the end of the run
4.
A fly wheel is speeded up from 5 rpm to 11 rpm in 100s.
The radius of the fly wheel is 0.08m. Calculate
a) the angular acceleration of the flywheel
b) the acceleration of a point on the rim along its circular path
Newton’s Laws of Motion
1.
A particle remains at rest or continues to move
in a straight line with uniform velocity unless
an unbalancing force act on it.
2.
The acceleration of a particle is proportional to
the resultant force acting on it and in the
direction of the force.
3.
The forces of action and reaction between
interacting bodies are equal in magnitude,
opposite in direction and collinear.
Momentum
 The momentum of a moving object is defined as:
Momentum = mass x velocity
 The principle of conservation of momentum states that for a
system of interaction objects, the total momentum remains
constant, provided no external resultant force acts on the
uA
uB
system.
Before
impart
During
impart
After
impart
A
B
A B
vA
vB
A
B
mAvA + mBvB = mAuA + mBuB
Exercise
1. A trolley of mass 2kg is moving at 5m/s. It collides
with a second, stationary, trolley of mass 8kg; it
bounces back with a velocity of 3m/s. With what
velocity does the second trolley move off?
2. Two billiard ball collide. Before the collision, ball A
is travelling at 1.5m/s towards stationary ball B.
After the collision, ball A travels in the same
direction but at 0.6m/s. If ball A has a mass of
0.3kg and ball B has a mass of 0.35kg, determine
the speed of ball B after impact.
Answer: 1. 2m/s 2. 0.77m/s
Exercise
3. A railway wagon of mass 1500kg travelling at a
speed of 2m/s, collides with three identical wagon
initially at rest. The wagons coupled together as a
result of the impact.
Calculate the
a) speed of the wagons after impact,
b) loss of kinetic energy due to the impact.
Answer:
a) 0.5m/s b) 2250J
Exercise
4. The 2000kg vehicle, travelling at 20m/s, collides with a vehicle of mass
2200kg, parked out of gear with the handbrake on, and the two vehicles lock
together.
Calculate the
a) momentum of the first vehicle before impact and the velocity of both
vehicles immediately after impact.
b) The handbrake exerts a braking force of 10kN on the vehicles. Calculate
the deceleration of the two vehicles and the distance travelled before
they come to rest.
c) Calculate the angular deceleration of the road wheels of the first vehicle.
The diameter of the vehicle’s road wheels is 72cm
Solution:
Exercise
5. A vehicle of mass 2000kg accelerates from rest, covering
400m in 28 seconds. Find the
a) average acceleration and the velocity of the vehicle at the
end of the run.
b) the momentum of the vehicle at the end of the run.
c) The diameter of the vehicle’s road wheels is 72cm.
Calculate the angular velocity of the road wheels at the
end of the run and the average angular acceleration.
Solution:
α = a/r = 1.02/0.36 = 2.83 rad/s2